4 Answers
4

The standard way to derive the formula for $\sinh^{-1}x$ goes like this:

Put $y = \sinh^{-1}x$ so that $x = \sinh y = \frac{e^y - e^{-y}}{2}$.

Rearrange this to get $2x = e^y - e^{-y}$, and hence $e^{2y} -2xe^y-1=0$, which is a quadratic equation in $e^y$. You then solve the quadratic and take logs (and take care with the $\pm$ sign you get with the roots of the quadratic).

So is my route entirely invalid? I am comfortable with your standard method but I thought this route would be better since it would be deriving it directly from Euler's formula instead of from an equation derived from Euler's formula.
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KainuiJan 1 '13 at 21:24

Your method looks like it is aiming to get the inverse function of $y=\sin x$ OK, but I don't see how your method is going to get the inverse function of $y = \sinh x$.
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Old JohnJan 1 '13 at 21:31